QUESTION:
You are given a 0-indexed string pattern of length n consisting of the characters ‘I’ meaning increasing and ‘D’ meaning decreasing.
A 0-indexed string num of length n + 1 is created using the following conditions:
num consists of the digits ‘1’ to ‘9’, where each digit is used at most once. If pattern[i] == ‘I’, then num[i] < num[i + 1]. If pattern[i] == ‘D’, then num[i] > num[i + 1]. Return the lexicographically smallest possible string num that meets the conditions.
Example 1:
Input: pattern = "IIIDIDDD"
Output: "123549876"
Explanation:
At indices 0, 1, 2, and 4 we must have that num[i] < num[i+1].
At indices 3, 5, 6, and 7 we must have that num[i] > num[i+1].
Some possible values of num are "245639871", "135749862", and "123849765".
It can be proven that "123549876" is the smallest possible num that meets the conditions.
Note that "123414321" is not possible because the digit '1' is used more than once.
Example 2:
Input: pattern = "DDD"
Output: "4321"
Explanation:
Some possible values of num are "9876", "7321", and "8742".
It can be proven that "4321" is the smallest possible num that meets the conditions.
Constraints:
1 <= pattern.length <= 8
pattern consists of only the letters 'I' and 'D'.
EXPLANATION:
虽然是medium的题目, 但是用贪心算法就可以解决出结果了. 如果要数字最小, 那么一定是最高位是最小的数字. 那么我们就可以从1开始进行. 如果是I就可以直接加入. 但是如果是D, 那么就需要反过来进行加入即可.
SOLUTION:
class Solution {
fun smallestNumber(pattern: String): String {
var sb:StringBuilder = StringBuilder()
var temp:StringBuilder = StringBuilder()
var ch:Char = '1'
pattern.forEach { c ->
temp.append(ch++)
if (c == 'I') {
sb.append(temp.reverse())
temp = StringBuilder()
}
}
temp.append(ch)
sb.append(temp.reverse())
return sb.toString()
}
}